On Minimum Dominating Sets in cubic and (claw,H)-free graphs

Given a graph G=(V,E), S⊆ V is a dominating set if every v∈ V∖ S is adjacent to an element of S. The Minimum Dominating Set problem asks for a dominating set with minimum cardinality. It is well known that its decision version is NP-complete even when G is a claw-free graph. We give a complexity dichotomy for the Minimum Dominating Set problem for the class of (claw, H)-free graphs when H has at most six vertices. In an intermediate step we show that the Minimum Dominating Set problem is NP-complete for cubic graphs.

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1 Introduction

We will only be concerned with simple undirected graphs. The reader is referred to [3] and [11] for, respectively, the definitions and notations on graph theory and on computational complexity.

Given a graph , a set is called a dominating set if every vertex is either an element of or is adjacent to an element of . When is a dominating set of we say that dominates . The minimum cardinality of a dominating set in is denoted by . A dominating set with is called a Minimum Dominating Set, a mds for short. Following [12] a mds is also called a -set.

Our aim is to determine the computational complexity of the task consisting of computing a -set or the domination number, for some subclasses of graphs defined by a finite set of forbidden subgraphs.

The decision problem associated with the Minimum Dominating Set is defined as:

.99 Minimum Dominating Set (MDS problem)

[2pt]     Instance: a graph and an integer . Question: is ?

In this paper we focus on the -free graphs complexity. The decision problems we study are defined this way.

.99 -free Minimum Dominating Set

[2pt]     Instance: a -free graph and an integer . Question: is ?

The paper is organized as follows. The two next sections give the notations, the results of the literature and some basic properties that will be used in the sequel of the paper. Then we prove that the MDS problem is -complete for the class of cubic graphs, a result that is strangely missing in the literature. This result and its proof will be useful for several demonstrations later on. The sections 5 to 8 are concerned with the MDS problem in the class of claw-free graphs when at least one other graph is excluded. Among our different results, in the section 7 we give a complexity dichotomy for the class of -free graphs for all the graphs with no more than six vertices. We give a partial result for the -free MDS problem when has at least seven vertices in the section 8. We summarize our main results and the problems left open in the conclusion.

2 Definitions and notations

An element is called an edge, if then is called a non-edge. For a vertex let us denote by its neighborhood, its closed neighborhood. The set of vertices at distance exactly of a vertex is denoted by . Hence and . A vertex is isolated if . A vertex is universal if . Two distinct vertices are twins if , are false twins if . The graph is the complementary graph of , that is and .

For a subset , we let denote the subgraph of induced by , which has vertex set  and edge set . Moreover, for a vertex , we write and for a subset we write . For a set of graphs, is -free if has no induced subgraph isomorphic to a graph in ; if we may write -free instead of -free. For two vertex disjoint induced subgraphs , is complete to if is an edge for any and , is anti-complete to if is an non-edge for any and . We denote by the disjoint union of the graphs and .

A set is called a stable set or an independent set if any pairwise distinct vertices are non adjacent. The maximum cardinality of an independent set in is denoted by . A set is called a clique if any pairwise distinct vertices are adjacent. When is a clique then is a complete graph. We denote by the clique or the complete graph on vertices and is the disjoint union of cliques ().

For , the graph denotes the cordless path on vertices, that is, and . For , the graph denotes the cordless cycle on vertices, that is, and . For , is called a hole. The graph is also called the triangle. The claw is the 4-vertex star, that is, the graph with vertices , , , and edges , , . The diamond is the 4-vertex complete graph minus an edge. The net is the graph with six vertices , , , and edges . The bull is the graph with five vertices , , and edges . The paw is the graph with four vertices , and edges . The butterfly is the graph with five vertices and edges . The house is the graph with five vertices and edges . The gem is the graph with five vertices and edges , . The Figure 2 below exposed all these graphs.

Figure 1: The claw, the diamond, the paw, the bull, the net.
Figure 2: The butterfly, the house, the gem.

The -triangle consists of a triangle and three vertex disjoint paths connected to . Hence the net is the -triangle, the bull is the -triangle, the paw is the -triangle. The -double-triangle consists of two vertex disjoint triangles and between and . For simplicity the -double-triangle is called the double-triangle (see Figure 3).

Figure 3: The -triangle, the -double-triangle, the double-triangle.

We denote the subset of vertices of such that . A mds which is also an independent set is an independent dominating set and the minimum cardinality of an independent dominating set is denoted by . Clearly . Note also that a minimum independent dominating set is a minimum maximal independent set.

3 Preliminary results

We know from [1] that if the graph is -free then . From [14] the MDS problem is -complete for the clas of -free graphs.

The minimum edge dominating set problem consists of finding a minimum set of edges such that for each edge , is incident to an edge . Taking , the line graph of , a minimum edge dominating set in is a minimum dominating set in . In [14] Yannakakis et al. showed that the minimum edge dominating set problem is -complete for bipartite subcubic graphs. Moreover, one can check that the graphs built in the transformation (from a variant of the -SAT problem) are

-free. Also, line graphs of bipartite graphs are perfect, so they have no odd holes. Thus, for any of these graphs, the corresponding line graphs are

-free and perfect (i.e. -free). It follows that the minimum dominating set problem is -complete for -free perfect graphs.

The minimum dominating set problem is polynomial for -free graphs. The class of -free graphs has bounded clique-width [7], so from [9] a -set can be computed in linear time. It is also polynomial for -free graphs [6]. A minimum dominating set can be computed in polynomial time for -free graphs so for -free graphs [2]. In [4], it is shown that computing a minimum dominating set is polynomial for the class of -free graphs. Since and computing a mds is polynomial for the classes of -free and -free graphs.

Note that -free graphs is equivalent to -free graphs when contains a . Hence, the remaining ’s we consider are -free.

We give some preliminary easy properties that will be useful for many proofs given later.

Property 3.1

Let , two fixed graphs with . If -free Minimum Dominating Set is then -free Minimum Dominating Set is .

Property 3.2

Let , two fixed graphs with . If -free Minimum Dominating Set is then -free Minimum Dominating Set is .

Property 3.3

Let be a fixed positive integer and a graph. If there exists of size such that then computing a minimum dominating set for is polynomial.

Proof:  Since and , we have that . So a minimum dominating set can be computed in which is polynomial.   

Property 3.4

Let be any fixed positive integer. If a graph is -free then computing a minimum dominating set for is polynomial.

Proof:  Since is -free we have . Since is claw-free . Now is fixed, so enumerating all the independent sets of size less than can be done in polynomial time. It follows that for any fixed , -free Minimum Dominating Set is polynomial.   

Property 3.5

Let be a connected -free graph with such that is a leaf. The Minimum Dominating Set problem is polynomial for if and only if it is polynomial for .

Proof:  Since is -free is a clique. Let , we show that . For contradiction we assume that . As shown in [5], is in every -set of . Let be a -set with . Let . If then is a dominating set, a contradiction. It follows and since is -free, is a clique. Let . is another -set, a contradiction. Hence . Since consists of and the component , we have that . Then from a -set of we obtain a -set of in polynomial time. Reciprocally, let be a -set of . Since is a leaf we assume that . Then is a -set for . Trivially it can be done in polynomial time from .   

4 Cubic and -regular graphs for when is odd

In [10] Demange et al. it is proved that, for any fixed , the Minimum Edge Dominating Set problem is -complete for bipartite -regular graphs. Transferring this result into the line graph we have that the Minimum Dominating Set problem is -complete for -regular graphs, , that is regular graphs with even degree at least four.

Theorem 4.1

The Minimum Dominating Set problem is -complete for cubic graphs.

Proof:  We give a polynomial reduction from the Minimum Dominating Set problem which is -complete for -regular graphs.

Let be a -regular graph and be an instance of Minimum Dominating Set. We build and another instance of Minimum Dominating Set where is cubic. Let be the number of vertices of , we take . Each -vertex of is replaced with the gadget shown in Figure 4. We denote by the gadget associated with .

The gadget has four corners, that is the vertices in Figure 4. Each of them corresponds to an edge incident to in . To the edge in corresponds an edge connecting the two corners of and . Hence is cubic.

For each , the subgraph of induced by the vertices of , satisfies the following properties. Note that by symmetry the four corners play the same role. Clearly and there exists a minimum dominating set that contains the four corners, see Figure 5. For each corner, says , has a unique minimum dominating set of size , see Figure 6. None of the three other corners are in this minimum dominating set. With ten vertices it is not possible to dominate .

Figure 4: A -vertex and its associated gadget .

Let be a dominating set of such that . We give a dominating set of as follows: if then the twelve vertices of the minimum dominating set of (the black vertices of Figure 5), are put in . If then is dominated by a vertex . Let . The eleven vertices of the minimum dominating set of (the black vertices of Figure 6), are put in . Hence the corner of is dominated by the corner of . So is a dominating set of size .

Figure 5: A minimum dominating set for the gadget.
Figure 6: A partial dominating set of size for the gadget.

Let be a dominating set of such that . We show that . From the last property, we gave at least eleven vertices to dominate each (note that it is necessary to dominate ). Since at most gadgets must contain at least vertices. Since twelve vertices are enough for one , we can assume there are gadgets with . Taking the corresponding vertices in , is a dominating set of . Hence we have that .   

The graphs built in the proof above are -free, so we have the following.

Corollary 4.2

The Minimum Dominating Set problem is -complete for -free cubic graphs.

Theorem 4.3

For any odd , the Minimum Dominating Set problem is -complete for -regular graphs.

Proof:

Figure 7: A -vertex and its replacement in the case of -regular graphs. The bold edge means that is connected to the vertices of .

The case is proved by Theorem 4.1. Now let be an odd integer. Given a cubic graph , we construct a graph as follows. Each vertex is connected to components (the complete graph with vertices minus an edge). For each component , let . The vertex is connected to and . Figure 7 shows the transformation for . For each component, one vertex is necessary in a minimum dominating set. Since is not dominated by , the sequel of the proof is easy.   

5 -free graphs

With similar reduction as the one used for proving Theorem 4.1, we show the following theorem.

Theorem 5.1

For any fixed integer the Minimum Dominating Set problem is -complete for -free subcubic graphs.

Proof:  The arguments are similar of those given in the proof of Theorem 4.1.

We give a polynomial reduction from the Minimum Dominating Set problem which is -complete for -regular graphs. To each -vertex is associated the gadget depicted in Figure 8. In this gadget each dashed box corresponds to an induced path of vertices, .

Figure 8: A -vertex and its associated gadget .

We have the following: and there exists a minimum dominating set that contains the four corners. For each corner, says , has a unique minimum dominating set of size . None of the three other corners are in this minimum dominating set. With vertices it is not possible to dominate .

The graph we obtain has no claw. Its maximum degree is three. The smallest cycle that is not a triangle has length more than . As can be chosen arbitrary large taking we obtain the result.   

6 -free graphs

Using the Theorem 4.1, we show the -completeness of the MDS problem for the class of -free subcubic graphs. In a first step we prove the following lemma.

Lemma 6.1

The -free Minimum Dominating Set problem is -complete for cubic perfect graphs.

Proof:  We give a polynomial reduction from the Minimum Dominating Set problem for cubic graphs which is proved -complete in Theorem 4.1.

Let be a cubic graph and be an instance of Minimum Dominating Set. We build and another instance of Minimum Dominating Set where is -free and perfect. Let be the number of vertices in , we take . Each -vertex of is replaced with the gadget shown in Figure 9. We denote by the gadget associated with .

The gadget has three corners - the vertices in Figure 9 - each of them corresponding to an edge incident to in . To the edge in corresponds an edge connecting the two corners of and . Hence is , , , -free. Moreover has no odd hole. Since is isomorphic to we have . One can remark that for any we have . Thus being -free, has no odd anti-hole. Then from the perfect graph theorem [8] is perfect.

For each , the subgraph of induced by the nine vertices of , satisfies the following properties. By symmetry the three corners play the same role. Clearly and there exists a minimum dominating set that contains the three corners, see Figure 10. For each corner, says , has a unique minimum dominating set of size , see Figure 10. None of the two other corners are in this minimum dominating set. With one vertex it is not possible to dominate .

Figure 9: A -vertex and its associated gadget .
Figure 10: A minimum dominating set and partial dominating set of size for the gadget

Let be a dominating set of such that . We give a dominating set of as follows: If then the three vertices of the minimum dominating set of (the three black vertices of Figure 10), are put in . If then is dominated by a vertex . Let . The two vertices of the minimum dominating set of (the two black vertices of Figure 10), are put in . Hence the corner of is dominated by the corner of . So is a dominating set of size .

Let be a dominating set of such that . We show that . From the last property we gave for at least two vertices are necessary to dominate . Since at most gadgets must contain at least three vertices. Since three vertices are enough for one , we can assume there are gadgets with . Taking the corresponding vertices in , is a dominating set of . Hence we have that .   

So we can prove the following theorem.

Theorem 6.2

For any fixed integer the Minimum Dominating Set problem is -complete for -free subcubic graphs.

Proof:  The arguments are similar of those given in the proof of Lemma 6.1.

We give a polynomial reduction from the Minimum Dominating Set problem which is -complete for cubic graphs. Let be a cubic graph. To each -vertex is associated the gadget depicted by Figure 11. In this gadget each dashed box corresponds to an induced path of vertices (we depicted the case ). To each edge is associated a (the same dashed box in the picture). We take .

Figure 11: A -vertex and its associated gadget .
Figure 12: On the left : the vertex is in the dominating set; on the right: the vertex is not in the dominating set.

The graph we obtain is -free and its maximum degree is three. As shown by Figure 12 the remaining arguments are the same as in the proof of Lemma 6.1.

As is a positive integer that can be chosen arbitrary large we obtain the result.   

7 graphs

For any fixed graph with no more than six vertices we determine the complexity of the MDS problem for the class of -free graphs. Trivially the graphs we consider are assumed to be connected. When has no more than two vertices the MDS problem is trivially polynomial, so we prove the results for all the graphs with three vertices up to six vertices.

Property 7.1

For any fixed graph with three vertices computing a minimum dominating set for a connected -free graph is polynomial.

Proof:  Since a connected -free graph is either a path or a cycle, the MDS problem is polynomial when is -free with . When we have . From [4] we know that computing a mds is polynomial for -free graphs. Then the result holds from Property 3.2.   

The next three sections deal with the case where contains four, five, six vertices.

7.1 graphs when has four vertices

When has exactly four vertices the state of the art is the following. In [14], the minimum edge dominating set problem is shown to be -complete for bipartite subcubic graphs. Moreover, one can check that the graphs built in the transformation (from a variant of the -SAT problem) are -free. Also, line graphs of bipartite graphs are perfect, so they have no odd holes. Thus, the line graph of such a is -free and perfect (i.e., -free). It follows that the minimum dominating set problem is -complete for -free perfect graphs. From [4] computing a mds is polynomial for -free graphs when . It is also polynomial when since . The class of -free graphs has bounded clique-width [7], so from [9] a -set can be computed in linear time (remark that .

Taking these results together we obtain the following dichotomy.

Property 7.2

Let be a fixed graph with four vertices. Computing a minimum dominating set for a connected -free graph is -complete when , otherwise it is polynomial.

Remark 7.1

In [13] Lin et al. remark that the minimum dominating set problem is -complete for -free graphs, here we restrict this result for the class of perfect graphs.

7.2 graphs when has five vertices

In this section, we focus on with exactly five vertices. From [14], we know that the MDS problem is -complete when is -free and perfect, hence for -free graphs. Also, from Lemmas 3.1 and 6.1, we know that the MDS problem is -complete for -free graphs when contains a , a , a , or a .

Using Property 3.1 the MDS problem is NP-complete when . From Property 3.2, it is polynomial for -free graphs when or , and from Property 3.4 it is polynomial for -free graphs. Hence, the MDS problem is polynomial for .

It remains to treat the cases where .

Property 7.3

Let be a connected -free graph. If then computing a minimum dominating set for is polynomial.

Proof:  When is a -free graph computing a mds is polynomial. Now we suppose that . Let . From Property 3.3 we can assume that . Let . If is not complete then there are such that , but . Thus is complete and (taking and ). Hence a mds can be computed in time .  

Property 7.4

Let be a connected -free graph. If then computing a minimum dominating set for is polynomial.

Proof:  When is a -free graph computing a mds is polynomial (see [4]). Now we suppose that . Let . From Property 3.3 we can assume that . Let . Since is connected and -free, there are such that is a neighbor of and has exactly two neighbors in . By symmetry we can assume that . Clearly , and .   

Taking the results together we obtain the following dichotomy.

Theorem 7.1

Let be a fixed graph with five vertices. Computing a minimum dominating set for a connected -free graph is -complete when and polynomial otherwise.

7.3 graphs when has six vertices

We consider with exactly six vertices. From Theorems 5.1, 6.2, the MDS problem is -complete when is -free (for any fixed ) and when is -free (for any fixed ). Hence, the MDS problem is -complete for -free graphs and for -free graphs. Furthermore, the MDS problem is -complete for -free graphs when contains at least one of the following graphs: .

Also from lemma 3.2, the MDS problem is for -free graphs when and the MDS problem is for -free graphs. Hence, if is a net or is an induced subgraph of , then the MDS problem is for -free graphs.

So, we focus on -free graphs where the complexity of the MDS problem cannot be deduced from the previous arguments.

Property 7.5

Let a connected -free graph. Computing a minimum dominating set for is polynomial if .

Proof:  Let a maximum induced path of . Since computing a minimum dominating set is polynomial for -free graphs [4], we can assume that . Since is -free, any vertex is such that . Also, from Property 3.3, we can assume that , so . Let with a neighbor such that . Since is claw-free has exactly two neighbors that are consecutive in , that is , . Since is a maximal path, these two neighbors cannot be or . Hence , .

From Property 3.5 we can assume that both and are not leaves. Let be two neighbors of and respectively. We show how the vertices and are connected to . First, we deal with . From above has no neighbor in . If or then has a claw. Hence, when , and when , or . Last, if , i.e. then , a contradiction. Hence there is no vertex such that . Now when we have or or , and or